Fdtxjr

Why no variance term in Bayesian logistic regression?

Is "remove commented out code" correct English?

Method Does Not Exist error message

Are there any examples of a variable being normally distributed that is *not* due to the Central Limit Theorem?

Why is it a bad idea to hire a hitman to eliminate most corrupt politicians?

How do conventional missiles fly?

Should I cover my bicycle overnight while bikepacking?

Why would the Red Woman birth a shadow if she worshipped the Lord of the Light?

What mechanic is there to disable a threat instead of killing it?

How do I handle a potential work/personal life conflict as the manager of one of my friends?

What's the in-universe reasoning behind sorcerers needing material components?

How seriously should I take size and weight limits of hand luggage?

Could the museum Saturn V's be refitted for one more flight?

Forming a German sentence with/without the verb at the end

Can I run a new neutral wire to repair a broken circuit?

How to show a landlord what we have in savings?

How can saying a song's name be a copyright violation?

Why was the shrinking from 8″ made only to 5.25″ and not smaller (4″ or less)?

Ambiguity in the definition of entropy

Is this a hacking script in function.php?

One verb to replace 'be a member of' a club

Is it possible to create a QR code using text?

Valid term from quadratic sequence?

Avoiding the "not like other girls" trope?

Linear transformation with change of ordered basis

How do I find transformation matrix with respect to given basis in the domain and/or the codomain, given the transformation in the standard basis?I need help with this linear transformation.Linear transformation: Change of basisLinear Algebra Change of Basis problemLinear transformation change of basisFind the matrix of linear transformation with respect to different basisFind the matrix of a linear transformationLinear Transformation Matrix with Change of BasisFind matrix of polynomial linear transformation relative to basisLinear transformation: change of basis matrix representation

0

$begingroup$

Let the linear transformation $ D: P_3 rightarrow P_2 $ be defined as $ D(f(x)) = f'(x) $.

$
B = lbrace 1, x, x^2, x^3 rbrace, \
C = lbrace 2+x+x^2, -2-x^2, 1-x rbrace
$

Find the matrix $ [D]_B^C $ for $ D $ relative to the basis $ B $ in the domain and $ C $ in the codomain.

---

So $ [D]_B^P_2 =
beginbmatrix
0&1&0&0\
0&0&2&0\
0&0&0&3
endbmatrix
$

and

$ T_P_2 rightarrow C = Tleft( beginbmatrix1\x\x^2\endbmatrix right) =
beginbmatrix2+x+x^2\-2-x^2\1-xendbmatrix$,
$
[T]_P_2^C=
beginbmatrix
2&1&1\
-2&0&-1\
1&-1&0
endbmatrix
$

---

$
T(D(x)) = [T]_P_2^C cdot [D]_B^P_2 =
beginbmatrix
0&2&2&3\
0&-2&0&-3\
0&1&-2&0
endbmatrix
$

Why doesn't this work?

linear-algebra linear-transformations

share|cite|improve this question

edited Mar 21 at 4:26

asdffdsa

asked Mar 21 at 4:21

asdffdsaasdffdsa

324

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand what you're doing here. What does $[D]_B^P_2$ represent? I thought $P_2$ was the space of polynomials of degree at most $2$, not a basis. Also, what is $T$?
  $endgroup$
  – Theo Bendit
  Mar 21 at 5:01
  

  
  
  
  

add a comment | 

0

$begingroup$

Let the linear transformation $ D: P_3 rightarrow P_2 $ be defined as $ D(f(x)) = f'(x) $.

$
B = lbrace 1, x, x^2, x^3 rbrace, \
C = lbrace 2+x+x^2, -2-x^2, 1-x rbrace
$

Find the matrix $ [D]_B^C $ for $ D $ relative to the basis $ B $ in the domain and $ C $ in the codomain.

---

So $ [D]_B^P_2 =
beginbmatrix
0&1&0&0\
0&0&2&0\
0&0&0&3
endbmatrix
$

and

$ T_P_2 rightarrow C = Tleft( beginbmatrix1\x\x^2\endbmatrix right) =
beginbmatrix2+x+x^2\-2-x^2\1-xendbmatrix$,
$
[T]_P_2^C=
beginbmatrix
2&1&1\
-2&0&-1\
1&-1&0
endbmatrix
$

---

$
T(D(x)) = [T]_P_2^C cdot [D]_B^P_2 =
beginbmatrix
0&2&2&3\
0&-2&0&-3\
0&1&-2&0
endbmatrix
$

Why doesn't this work?

linear-algebra linear-transformations

share|cite|improve this question

edited Mar 21 at 4:26

asdffdsa

asked Mar 21 at 4:21

asdffdsaasdffdsa

324

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand what you're doing here. What does $[D]_B^P_2$ represent? I thought $P_2$ was the space of polynomials of degree at most $2$, not a basis. Also, what is $T$?
  $endgroup$
  – Theo Bendit
  Mar 21 at 5:01
  

  
  
  
  

add a comment | 

$begingroup$

Let the linear transformation $ D: P_3 rightarrow P_2 $ be defined as $ D(f(x)) = f'(x) $.

$
B = lbrace 1, x, x^2, x^3 rbrace, \
C = lbrace 2+x+x^2, -2-x^2, 1-x rbrace
$

Find the matrix $ [D]_B^C $ for $ D $ relative to the basis $ B $ in the domain and $ C $ in the codomain.

---

So $ [D]_B^P_2 =
beginbmatrix
0&1&0&0\
0&0&2&0\
0&0&0&3
endbmatrix
$

and

$ T_P_2 rightarrow C = Tleft( beginbmatrix1\x\x^2\endbmatrix right) =
beginbmatrix2+x+x^2\-2-x^2\1-xendbmatrix$,
$
[T]_P_2^C=
beginbmatrix
2&1&1\
-2&0&-1\
1&-1&0
endbmatrix
$

---

$
T(D(x)) = [T]_P_2^C cdot [D]_B^P_2 =
beginbmatrix
0&2&2&3\
0&-2&0&-3\
0&1&-2&0
endbmatrix
$

Why doesn't this work?

linear-algebra linear-transformations

share|cite|improve this question

edited Mar 21 at 4:26

asdffdsa

asked Mar 21 at 4:21

asdffdsaasdffdsa

324

$endgroup$

share|cite|improve this question

edited Mar 21 at 4:26

asdffdsa

asked Mar 21 at 4:21

asdffdsaasdffdsa

324

share|cite|improve this question

edited Mar 21 at 4:26

asdffdsa

asked Mar 21 at 4:21

asdffdsaasdffdsa

324

asked Mar 21 at 4:21

asdffdsaasdffdsa

324

asked Mar 21 at 4:21

asdffdsaasdffdsa

324

1 Answer
1

active

oldest

votes

1

$begingroup$

The matrix from basis $C$ to $B$ is given by $$T_Cto P_2=beginpmatrix2&-2&1\1&0&-1\1&-1&0endpmatrix$$ because if you take, say, the $(1,0,0)$ vector, it is mapped to $(2,-2,1)$ in terms of the basis $B$. So the matrix you require is $$(T_Cto P_2)^-1D_B^P_2=beginpmatrix0&1&2&-6\0&1&2&-9\0&1&0&-6endpmatrix$$

share|cite|improve this answer

answered Mar 21 at 12:57

ChrystomathChrystomath

1,978513

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3156328%2flinear-transformation-with-change-of-ordered-basis%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

The matrix from basis $C$ to $B$ is given by $$T_Cto P_2=beginpmatrix2&-2&1\1&0&-1\1&-1&0endpmatrix$$ because if you take, say, the $(1,0,0)$ vector, it is mapped to $(2,-2,1)$ in terms of the basis $B$. So the matrix you require is $$(T_Cto P_2)^-1D_B^P_2=beginpmatrix0&1&2&-6\0&1&2&-9\0&1&0&-6endpmatrix$$

share|cite|improve this answer

answered Mar 21 at 12:57

ChrystomathChrystomath

1,978513

$endgroup$

add a comment | 

1

$begingroup$

The matrix from basis $C$ to $B$ is given by $$T_Cto P_2=beginpmatrix2&-2&1\1&0&-1\1&-1&0endpmatrix$$ because if you take, say, the $(1,0,0)$ vector, it is mapped to $(2,-2,1)$ in terms of the basis $B$. So the matrix you require is $$(T_Cto P_2)^-1D_B^P_2=beginpmatrix0&1&2&-6\0&1&2&-9\0&1&0&-6endpmatrix$$

share|cite|improve this answer

answered Mar 21 at 12:57

ChrystomathChrystomath

1,978513

$endgroup$

add a comment | 

$begingroup$

The matrix from basis $C$ to $B$ is given by $$T_Cto P_2=beginpmatrix2&-2&1\1&0&-1\1&-1&0endpmatrix$$ because if you take, say, the $(1,0,0)$ vector, it is mapped to $(2,-2,1)$ in terms of the basis $B$. So the matrix you require is $$(T_Cto P_2)^-1D_B^P_2=beginpmatrix0&1&2&-6\0&1&2&-9\0&1&0&-6endpmatrix$$

share|cite|improve this answer

answered Mar 21 at 12:57

ChrystomathChrystomath

1,978513

$endgroup$

share|cite|improve this answer

answered Mar 21 at 12:57

ChrystomathChrystomath

1,978513

answered Mar 21 at 12:57

ChrystomathChrystomath

1,978513

answered Mar 21 at 12:57

ChrystomathChrystomath

1,978513